Question

Find the equation of the plane passing through the points \((-2i + 6j - 6k), (-3i + 10j - 9k), (-5i - 6j - 6k)\).

Hint:

For finding the equation of a plane passing through three points, use the cross product to find the normal vector and apply the point-normal form of the plane equation.

Answer 1

Step 1: Write the coordinates of the points.
Let the points be \( A(-2, 6, -6) \), \( B(-3, 10, -9) \), and \( C(-5, -6, -6) \).
Step 2: Find two vectors in the plane.
\[ \overrightarrow{AB} = B - A = (-3 + 2, 10 - 6, -9 + 6) = (-1, 4, -3) \] \[ \overrightarrow{AC} = C - A = (-5 + 2, -6 - 6, -6 + 6) = (-3, -12, 0) \]
Step 3: Find the normal vector to the plane.
The normal vector is the cross product of \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \): \[ \overrightarrow{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -12 & 0 \end{vmatrix} \] \[ \overrightarrow{n} = \hat{i} \left( 4(0) - (-3)(-12) \right) - \hat{j} \left( -1(0) - (-3)(-3) \right) + \hat{k} \left( -1(-12) - 4(-3) \right) \] \[ \overrightarrow{n} = \hat{i} (0 - 36) - \hat{j} (0 - 9) + \hat{k} (12 - (-12)) = -36\hat{i} + 9\hat{j} + 24\hat{k} \] Thus, the normal vector is \( \overrightarrow{n} = (-36, 9, 24) \).
Step 4: Find the equation of the plane.
The equation of the plane is: \[ -36(x + 2) + 9(y - 6) + 24(z + 6) = 0 \] Simplifying: \[ -36x - 72 + 9y - 54 + 24z + 144 = 0 \] \[ -36x + 9y + 24z + 18 = 0 \]
Final Answer:
\[ \boxed{-36x + 9y + 24z + 18 = 0} \]